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If A+B+C=\pi and \frac{\sin 2 A+\sin 2 B+\sin 2 C}{\sin A+\sin B+\sin C}  \lambda \sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right), then the value of 1+2 \lambda+3 \lambda^2+4 \lambda^3 must be 

Option: 1

2000


Option: 2

2257


Option: 3

2256

 


Option: 4

2210


Answers (1)

best_answer

\because \sin 2 A+\sin 2 B+\sin 2 C=4 \sin A \sin B \sin C

          =32 \sin A / 2 \sin B / 2 \sin C / 2 \cos A / 2 \cos B / 2 \cos C / 2

and  \sin A+\sin B+\sin C=4 \cos A / 2 \cos B / 2 \cos C / 2

                                                                        (from conditional identities)

\therefore \frac{\sin 2 A+\sin 2 B+\sin 2 C}{\sin A+\sin B+\sin C}=8 \sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right)

On comparing, we get \lambda=8

Then, 1+2 \lambda+3 \lambda^2+4 \lambda^3

                                \begin{aligned} & =1+16+3(64)+4(512) \\ \\& =2257 \end{aligned}

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shivangi.shekhar

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